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In this post we present the solution with Maple to the logical problem of "Gardens Puzzle"

http://www.mathsisfun.com/puzzles/gardens-solution.html

The Puzzle:

Five friends have their gardens next to one another, where they grow three kinds of crops: fruits (apple, pear, nut, cherry), vegetables (carrot, parsley, gourd, onion) and flowers (aster, rose, tulip, lily).

1. They grow 12 different varieties.
2. Everybody grows exactly 4 different varieties
3. Each variety is at least in one garden.
4. Only one variety is in 4 gardens.
5. Only in one garden are all 3 kinds of crops.
6. Only in one garden are all 4 varieties of one kind of crops.
7. Pear is only in the two border gardens.
8. Paul's garden is in the middle with no lily.
9. Aster grower doesn't grow vegetables.
10. Rose growers don't grow parsley.
11. Nuts grower has also gourd and parsley.
12. In the first garden are apples and cherries.
13. Only in two gardens are cherries.
14. Sam has onions and cherries.
15. Luke grows exactly two kinds of fruit.
16. Tulip is only in two gardens.
17. Apple is in a single garden.
18. Only in one garden next to Zick's is parsley.
19. Sam's garden is not on the border.
20. Hank grows neither vegetables nor asters.
21. Paul has exactly three kinds of vegetable.

Who has which garden and what is grown where?

 

About methods of solution. At first I just wanted to generate all variations and using conditions 1 .. 21 to find all solutions. But even if we use the condition that everybody grows exactly 4 different varieties then the total number variants equals  5!^2*binomial(12,4)^5=427945522455000000

So from the very beginning using some of the conditions 1 .. 21 we maximally reduce the number of possible variants. For example from the conditions 11, 18 and 6 implies that only in one garden are all 4 varieties of flowers. Next we pass through these variants and using conditions 1 .. 21 and finally come to a unique solution:

restart;

Fruits:={apple, pear, nut, cherry}:

Vegetables:={carrot, parsley, gourd, onion}:

Flowers:={aster, rose, tulip, lily}:

 

Set1:=Flowers:

Garden1:={Set1}:

Set2:=Fruits union Vegetables union Flowers minus {nut,gourd,parsley} minus {apple,cherry,rose}:

Garden2:={seq({nut,gourd,parsley} union {Set2[i]}, i=1..nops(Set2))}:

Set3:=Vegetables union Flowers minus {parsley}:

Garden3:={seq({apple,cherry,pear} union {Set3[i]}, i=1..nops(Set3))}:

Set4:=combinat[choose](Fruits union Vegetables union Flowers minus {onion,cherry} minus {apple,parsley,pear,nut}, 2):

Garden4:={seq({onion,cherry} union Set4[i], i=1..nops(Set4))}:

Set5:=Fruits union Vegetables union Flowers minus {apple, cherry,nut,parsley}:

Garden5:=combinat[choose](Set5, 4):

 

S:=[]:

for s1 in Garden1 do

for s2 in Garden2 do

for s3 in Garden3 do

for s4 in Garden4 do

for s5 in Garden5 do

s:=[s1,s2,s3,s4,s5]: s_4:=combinat[choose](s,4): m:=0: n:=0: k:=0: p:=0: q:=0:

for i in s do

if `intersect`(i,Fruits)<>{} and `intersect`(i,Vegetables)<>{} and `intersect`(i,Flowers)<>{} then m:=m+1: fi:

if pear in i then n:=n+1: fi:

if tulip in i then k:=k+1: fi:

if aster in i and `intersect`(i,Vegetables)<>{} then p:=p+1: fi:

if i=Fruits or i=Vegetables or i=Flowers then q:=q+1: fi:

od:

if nops(`union`(op(s)))=12 and nops(`union`(seq(`intersect`(op(s_4[j])), j=1..nops(s_4))))=1 and m=1 and n=2 and k=2 and p=0 and q=1 then S:=[op(S),[s1,s2,s3,s4,s5]]: fi:

od: od: od: od: od:

 

L1:=[seq([[3,Paul],[2,Sam],seq([combinat[permute]([1,4,5])[i,j],[Luke,Zick,Hank][j]],j=1..3)],i=1..6)]:

L2:=[seq([[3,Paul],[4,Sam],seq([combinat[permute]([1,2,5])[i,j],[Luke,Zick,Hank][j]],j=1..3)],i=1..6)]:

L0:=[op(L1),op(L2)]:

L:=[seq(op(combinat[permute](L0[i])),i=1..nops(L0))]:

Sol:=[]:

for l in L do

for s in S do

sol:=[seq([op(l[i]),s[i]], i=1..5)]:

gad1:=op(select(has,sol,parsley)): gad2:=op(select(has,sol,Zick)):

if abs(gad1[1]-gad2[1])=1 and convert([seq(((pear in sol[i][3] implies (sol[i][1]=1 or sol[i][1]=5)) and (sol[i][2]=Paul implies (not lily in sol[i][3])) and (sol[i][1]=1 implies (apple in sol[i][3] and cherry in sol[i][3])) and (sol[i][2]=Sam implies (onion in sol[i][3] and cherry in sol[i][3])) and (sol[i][2]=Luke implies nops(`intersect`(sol[i][3],Fruits))=2) and (sol[i][2]=Hank implies (`intersect`(sol[i][3],Vegetables)={} and not aster in sol[i][3])) and (sol[i][2]=Paul implies nops(`intersect`(sol[i][3],Vegetables))=3)), i=1..5)], `and`) then Sol:=[op(Sol), sol]: fi:

 

od: od:

for i in Sol do

Matrix(sort(i,(x,y)->x[1]<y[1]));

od;

 

 

For some time I have been running into the problem where uploaded worksheets that are inlined into a Mapleprimes post get gridlines shown, even when they are not specified or shown in the original worksheet. This is a problem because quite often the gridlines make an inlined plot unattractive.

I believe that Mapleprimes uses some version of MapleNet as a mechanism for inlining uploaded worksheets.

It seems that the Standard GUI and/or MapleNet have special handling for gridlines in the sense that they actually distinguish between PLOT structures generated without the `gridlines` option versus those created with `gridlines=false`.  I believe that this has something to do with the persistence of various plot qualities in an output region which (perhaps for historical reasons?)  allow re-execution of a plot input command to produce a plot output rendering that remembers whether gridlines are visible.

If the second plot below does indeed get rendered here in Primes without gridlines visible then one workaround is evident: adding `gridlines=false` instead of omitting the option.


plot(sin(x),x=-Pi..Pi);

plot(sin(x),x=-Pi..Pi,gridlines=false);

 


Download test.mw

I'll submit a bug report against MapleNet.

acer

In this post we present another compact proof of this remarkable theorem without using  geometry package.
The proof uses a procedure called  Cc , which for three points returns a list of the coordinates of the center and the radius of the circumscribed circle.

restart;

Cc:=proc(A,B,C)

local x1, y1, x2, y2, x3, y3, x, y;

x1,y1:=op(A);  x2,y2:=op(B);  x3,y3:=op(C);

solve({(x2-x1)*(x-(x1+x2)/2)+(y2-y1)*(y-(y1+y2)/2)=0, (x2-x3)*(x-(x2+x3)/2)+(y2-y3)*(y-(y2+y3)/2)=0},{x,y});

assign(%);

[simplify([x,y]), simplify(sqrt((x-x1)^2+(y-y1)^2))];

end proc:

Proof for arbitrary triangle:

A, B, C:=[x1,y1], [x2,y2], [x3,y3]:

A1, B1, C1, M:=(B+C)/2, (A+C)/2, (A+B)/2, (A+B+C)/3:

P1:=Cc(A,M,B1)[1]: P2:=Cc(B1,M,C)[1]: P3:=Cc(C,M,A1)[1]:

P4:=Cc(A1,M,B)[1]: P5:=Cc(B,M,C1)[1]: P6:=Cc(C1,M,A)[1]:

Cc1:=Cc(P1,P2,P3):  Cc2:=Cc(P4,P5,P6):

is(Cc1=Cc2);

                                                  true

MapleSim 6.4 includes more powerful tools for creating custom components, performance enhancements, and enhancements to the model generators for Simulink® and FMI. 

We have also made important updates to the MapleSim Control Design Toolbox. This toolbox now offers a more complete set of algorithms for PID control, new commands for computing closed-loop transfer functions, and numerous improvements to existing commands. These enhancements allow engineers to design a greater variety of controllers and controller-observer systems while taking advantage of the greater flexibility and analysis options available through the use of symbolic parameters. 

See What’s New in MapleSim 6.4 and What’s New in the MapleSim Control Design Toolbox for details.

 

eithne

Let  us consider the general case of symbolic values C(xC,yC). I make use of the idea suggested by edgar in http://www.mapleprimes.com/questions/97743-How-To-Prove-Morleys-Trisector-Theorem : no assumptions.

restart; with(geometry); point(A, 0, 0);
point(B, 1, 0);
point(C, xC, yC);
point(MA, (xC+1)*(1/2), (1/2)*yC);
point(MC, 1/2, 0);
point(MB, (1/2)*xC, (1/2)*yC);
point(E, (0+1+xC)*(1/3), (0+0+yC)*(1/3));# the center of mass
line(l1, x = 1/4, [x, y]);
The coordinates of the center of the first described circle are found as the solutions of the system of the equations of midperpendiculars.

midpoint(ae, A, E); coordinates(ae);


S1 := solve({x = 1/4, ((xC+1)*(1/3))*(x-(xC+1)*(1/6))+(1/3)*yC*(y-(1/6)*yC) = 0}, {x, y});

BTW, Maple can't create the midperpendiculars in this case.

point(O1, op(map(rhs, S1)));
                               O1

Simple details are omitted in the above. The coordinates of the centers of the two next described circles are found similarly.
coordinates(midpoint(mce, MC, E));

S2 := solve({x = 3/4, ((-1/2+xC)*(1/3))*(x-5/12-(1/6)*xC)+(1/3)*yC*(y-(1/6)*yC) = 0}, {x, y});

point(O2, op(map(rhs, S2)));

                               O2
coordinates(midpoint(bma, B, MA)); coordinates(midpoint(be, B, E));
  

                

S3 := solve({(xC-1)*(x-(xC+3)*(1/4))+yC*(y-(1/4)*yC) = 0, ((-2+xC)*(1/3))*(x-(4+xC)*(1/6))+(1/3)*yC*(y-(1/6)*yC) = 0}, {x, y});

point(O3, op(map(rhs, S3)));

                               O3

Now we find the equation of the circumference which passes through O1, O2, and O3.

eq := a*x+b*y+x^2+y^2+c = 0:
sol := solve({eval(eq, S1), eval(eq, S2), eval(eq, S3)}, {a, b, c});

A long output can be seen in the attached .mw file.

eq1 := eval(eq, sol);

  Now we find (in suspense)  the coordinates of the next center and verify whether it belongs to the sircumference O1O2O3.

coordinates(midpoint(mac, C, MA)); coordinates(midpoint(ec, E, C)); S4 := solve({(xC-1)*(x-(3*xC+1)*(1/4))+yC*(y-3*yC*(1/4)) = 0, ((2*xC-1)*(1/3))*(x-(4*xC+1)*(1/6))+(2*yC*(1/3))*(y-4*yC*(1/6)) = 0}, {x, y});

 point(O4, op(map(rhs, S4)));

                               O4
simplify(eval(eq1, S4));

                             0 = 0

Hope the reader will have a real pleasure to find the two residuary centers and to verify these on his/her own.

geom2.mw

 

 

 

 

It is well known that the medians of a triangle divide it into 6 triangles.
It is less known that the centers of their circumscribed circles belong to one circumference as drawn below

This remarkable theorem  was proved in the 21st century! Unfortunately, I lost its source.
I can't prove this difficult  theorem by hand. However, I can prove it with Maple.
The aim of this post is to expose these proofs. Everybody knows that it is scarcely possible
to construct a general triangle with help of the geometry package of Maple.
Without loss of generality one may assume that the vertex A is placed at the origin,
the vertex B is placed at (1,0), and the vertex C(xC,yC). We firstly consider the theorem
in the case of concrete values of xC and yC.

restart; with(geometry):with(plots):
point(A, 0, 0);
point(B, 1, 0);
xC := 15*(1/10); yC := sqrt(3); point(C, xC, yC);
triangle(T, [A, B, C]);
median(mA, A, T, MA);
median(mB, B, T, MB);
median(mC, C, T, MC);
line(m1, [A, MA]);
line(m2, [B, MB]);
intersection(E, m1, m2);
triangle(AEMB, [A, E, MB]);
circumcircle(c1, AEMB, 'centername' = C1);
circumcircle(c2, triangle(CEMB, [C, E, MB]), 'centername' = C2);
circumcircle(c3, triangle(CEMA, [C, E, MA]), 'centername' = C3);
circumcircle(c4, triangle(BEMA, [B, E, MA]), 'centername' = C4);
circumcircle(c5, triangle(BEMC, [B, E, MC]), 'centername' = C5);
circumcircle(c6, triangle(AEMC, [A, E, MC]), 'centername' = C6);
circle(CC, [C1, C2, C3]);
IsOnCircle(C4, CC);
                              true

IsOnCircle(C5, CC);
                              true
IsOnCircle(C6, CC);
                              true
display([draw([T(color = black), mA(color = black), mB(color = black), mC(color = black), C1(color = blue), C2(color = blue), C3(color = blue), C4(color = blue), C5(color = blue), C6(color = blue), CC(color = red)], symbol = solidcircle, symbolsize = 15, thickness = 2, scaling = constrained), textplot({[-0.5e-1, 0.5e-1, "A"], [.95, 0.5e-1, "B"], [xC-0.5e-1, yC+0.5e-1, "C"]})], axes = frame, view = [-.1 .. max(1, xC)+.1, 0 .. yC+.1]);

This can be done as a procedure in such a way.

restart; SixPoints := proc (xC, yC) geometry:-point(A, 0, 0); geometry:-point(B, 1, 0); geometry:-point(C, xC, yC); geometry:-triangle(T, [A, B, C]); geometry:-median(mA, A, T, MA); geometry:-median(mB, B, T, MB); geometry:-median(mC, C, T, MC); geometry:-line(m1, [A, MA]); geometry:-line(m2, [B, MB]); geometry:-intersection(E, m1, m2); geometry:-triangle(AEMB, [A, E, MB]); geometry:-circumcircle(c1, AEMB, 'centername' = C1); geometry:-circumcircle(c2, geometry:-triangle(CEMB, [C, E, MB]), 'centername' = C2); geometry:-circumcircle(c3, geometry:-triangle(CEMA, [C, E, MA]), 'centername' = C3); geometry:-circumcircle(c4, geometry:-triangle(BEMA, [B, E, MA]), 'centername' = C4); geometry:-circumcircle(c5, geometry:-triangle(BEMC, [B, E, MC]), 'centername' = C5); geometry:-circumcircle(c6, geometry:-triangle(AEMC, [A, E, MC]), 'centername' = C6); geometry:-circle(CC, [C1, C2, C3]); return geometry:-IsOnCircle(C4, CC), geometry:-IsOnCircle(C5, CC), geometry:-IsOnCircle(C6, CC), geometry:-draw([CC(color = blue), C1(color = red), C2(color = red), C3(color = red), C4(color = red), C5(color = red), C6(color = red), T(color = black), mA(color = black), mB(color = black), mC(color = black), c1(color = green), c4(color = green), c2(color = green), c3(color = green), c5(color = green), c6(color = green)], symbol = solidcircle, symbolsize = 15, thickness = 2) end proc;
SixPoints(1.5, 1.2);

true, true, true, PLOT(...)
 SixPoints(1.5, 1.2)[4];

See geom1.mw

To be continued (The general case will be considered in  part 2http://www.mapleprimes.com/posts/200210-Six-Points-On-Circumference-2 .).

 

 





 

Maple 18, document mode. Please try this:

>restart;
>some_long_name:=3:
>f(som)

Now, as the cursor is right after the "m" above, hiting the ESC key to autocomplete, nothing happens. Only a beep, meaning Maple does not know about the variable name I wanted to expand to, which is "some_long_name". What I do now is manually add a space by  pushing the ")" away, like this:

  f(som )

And now put the mouse back in front of the "m" and now hit ESC. Maple now can see the name I wanted.

Really? This is very bad design. You might ask, why did I close the () first? Well, I like to start by writing () then fill it in, or it can be that I am changing things, and wanted to write a new name, and the () was allready there. I use name autocomplete alot, and it is very annoying to not be able to use it if happens that there is a ")" or "]" right next to the name.

Having to keep pushing the ")" away, so that Maple can see the name makes no sense since ")" can't part of a name.

In Mathematica it just works. I can do

someLongName = 3;

Then in a new cell,

f[som] and hit CTRL-K  when mouse at the letter "m" just typed it in, and it will autocomplete.  I did not have to push "]" away first.

Please Maple, fix your user interface so it is less awkward and annoying to use. This is version 18 of the software, and not version 1.

 

A post to conjure up some interest.  Nasa seeks coders to hunt asteroids http://www.bbc.com/news/technology-26528516

Just going out on a limb here but any of the three M's would have a good chance here.  A good opportunity to create some code.  Maple could port it to C code then compiled for speed.  On a smaller scale it would be an interesting challenge.  On the larger scale not sure if Maple would be best suited for the task.  Opinions?

This is one of my favorite events of the year. When we launch a new release of Maple, I get to see the work of so many talented individuals at Maplesoft come together in a form that I am sure will delight, and maybe even surprise you.

We are holding true to our principles with Maple 18. Hundreds of new mathematical algorithms further strengthen a computational engine that will help you tackle your toughest challenges. The user interface experience continues to become smarter, allowing you to focus on getting results without fighting with syntax. Connectivity options are again becoming richer.

A personal favorite of mine is the newly enhanced Explore functionality, which allows you to, with a couple of clicks, go from a mathematical expression to an interactive Math App. Math Apps allow you to explore the parameter space of the expression, gain insight into its behavior and even, in conjunction with Maple T.A., produce a gradeable Möbius App that allows you to assess a student’s interaction with the app and hence their understanding of the underlying concepts. The expanded Explore functionality is just part of a collection of advancements in Maple 18 that support The Möbius Project.

Overall, the new features of Maple 18 are quite numerous and I won’t try to list them all here. However, I do want to mention a few areas that have received special attention:

  • Statistics: Maple 18 includes lots of enhancements to statistics computations and visualization, such as new time series functionality that allows you to find patterns, make forecasts, and visualize time-based data. For the classroom, a new Student Statistics package, together with a range of bundled Math Apps, provide a simplified and interactive environment for instructors and students alike.

  • Physics: This package for representing and computing with concepts from general relativity to quantum mechanics continues to grow by leaps and bounds, with over 500 enhancements just in this release alone. We are convinced that this is the best computational environment available for researchers in this area.

  • Engineering: Key enhancements for control analysis, signal processing, and code generation to Python and Perl are just a few of the new features that engineers will note and appreciate. There’s even import/export for STL graphics files, which, amongst other things, means you can now print out your favorite Maple plots on a 3-D printer!

I think you will agree that Maple 18 exemplifies all the effort and attention that we have put into it.  And there’s more to come - this release is just the start of a stream of product announcements that you can expect from us in the coming months. Stay tuned!

I am using "Maple 15" and "Windows 7" and I want to share the tip for using "Classic worksheet" and "GR Tensor". Windows 7 is 64-bit but you can use Classic worksheet and GR Tensor, which work basically in 32-bit, by simply downlording "32-bit Maple 15" !! I haven't tried other more current versions of Maple but Maple 15 works well: you can choose 32-bit Maple in the download options.

Enjoy the Classic worksheet and GR Tensor. Good luck !  

On Thursday, Feb. 27, we are hosting our first-ever Virtual User Summit.   This day provides Maplesoft’s academic community a chance to learn more about the different ways Maplesoft technology is being used in education and research, a chance to interact with Maplesoft employees as well as each other, and a chance to get a glimpse into the future of education.

The virtual nature of this conference is a very tangible example of how much technology has changed our lives.  No less dramatic is the effect of technology on education.  In the keynote presentations at this conference, you will learn about Maplesoft’s vision for the future of education. You’ll also get to see tangible examples of technology that is building towards that vision, including sneak peeks of some things we are working on.

Visit Maplesoft Virtual User Summit for the full agenda and to register.  “Doors open” at 8:30 Eastern Time and the keynote presentations start at 9:00.

We are looking forward to this chance to come together and share our passion for technology and technical education.  Hope to see you there!

Vector using package Physics, LinearAlgebra.

Vectores.mw     (in spanish)

I was recently asked about performing some General Relativity computations from a paper by Plamen Fiziev, posted in the arXiv in 2013. It crossed my mind that this question is also instrumental to illustrate how these General Relativity algebraic computations can be performed using the Physics package. The pdf and mw links at the end show the same contents but with the Sections expanded.

General Relativity using Computer Algebra

 

Problem: for the spacetime metric,

g[mu, nu] = (Matrix(4, 4, {(1, 1) = -exp(lambda(r)), (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (2, 1) = 0, (2, 2) = -r^2, (2, 3) = 0, (2, 4) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = -r^2*sin(theta)^2, (3, 4) = 0, (4, 1) = 0, (4, 2) = 0, (4, 3) = 0, (4, 4) = exp(nu(r))}))

 

a) Compute the trace of

 

"Z[alpha]^(beta)=Phi R[alpha]^(beta)+`&Dscr;`[alpha]`&Dscr;`[]^(beta) Phi+T[alpha]^(beta)"

 

where `&equiv;`(Phi, Phi(r)) is some function of the radial coordinate, R[alpha, `~beta`] is the Ricci tensor, `&Dscr;`[alpha] is the covariant derivative operator and T[alpha, `~beta`] is the stress-energy tensor

 

T[alpha, beta] = (Matrix(4, 4, {(1, 1) = 8*exp(lambda(r))*Pi, (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (2, 1) = 0, (2, 2) = 8*r^2*Pi, (2, 3) = 0, (2, 4) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 8*r^2*sin(theta)^2*Pi, (3, 4) = 0, (4, 1) = 0, (4, 2) = 0, (4, 3) = 0, (4, 4) = 8*exp(nu(r))*Pi*epsilon}))

b) Compute the components of "W[alpha]^(beta)"" &equiv;"the traceless part of  "Z[alpha]^(beta)" of item a)

 

c) Compute an exact solution to the nonlinear system of differential equations conformed by the components of  "W[alpha]^(beta)" obtained in b)

 

Background: The equations of items a) and b) appear in a paper from February/2013, "Withholding Potentials, Absence of Ghosts and Relationship between Minimal Dilatonic Gravity and f(R) Theories", by Plamen Fiziev, a Maple user.  These equations model a problem in the context of a Branse-Dicke theory with vanishing parameter "omega." The Brans–Dicke theory is in many respects similar to Einstein's theory, but the gravitational "constant" is not actually presumed to be constant - it can vary from place to place and with time - and the gravitational interaction is mediated by a scalar field. Both Brans–Dicke's and Einstein's theory of general relativity are generally held to be in agreement with observation.

 

The computations below aim at illustrating how this type of computation can be performed using computer algebra, and so they focus only on the algebraic aspects, not the physical interpretation of the results.

a) The trace of "  Z[alpha]^(beta)=Phi R[alpha]^(beta)+`&Dscr;`[alpha]`&Dscr;`[]^(beta) Phi+T[alpha]^(beta)"

   

b) The components of "W[alpha]^(beta)"" &equiv;"the traceless part of " Z[alpha]^(beta)"

   

c) An exact solution for the nonlinear system of differential equations conformed by the components of  "W[alpha]^(beta)"

   

 

GeneralRelativit.pdf   GeneralRelativity.mw

Edgardo S. Cheb-Terrab 
Physics, Differential Equations and Mathematical Functions, Maplesoft

 

Code of the animation:

restart;

N := 192:

A := seq(plot([[.85*sin(t)^3-2+1.25*i/N, .85*(13*cos(t)*(1/15)-(1/3)*cos(2*t)-2*cos(3*t)*(1/15)-(1/15)*cos(4*t)), t = 0 .. Pi*i/N], [-.85*sin(t)^3-2+1.25*i/N, .85*(13*cos(t)*(1/15)-(1/3)*cos(2*t)-2*cos(3*t)*(1/15)-(1/15)*cos(4*t)), t = 0 .. Pi*i/N], [sin(t)^3+2-1.25*i/N, 13*cos(t)*(1/15)-(1/3)*cos(2*t)-2*cos(3*t)*(1/15)-(1/15)*cos(4*t), t = 0 .. Pi*i/N], [-sin(t)^3+2-1.25*i/N, 13*cos(t)*(1/15)-(1/3)*cos(2*t)-2*cos(3*t)*(1/15)-(1/15)*cos(4*t), t = 0 .. Pi*i/N]], color = red, thickness = 5, view = [-3 .. 3, -1.2 .. .9]), i = 1 .. N):

plots[display](A, insequence = true, scaling = constrained, axes = none);

Dear Bryon and Will:

Please fix MaplePrimes so that the Replies to Answers to Questions appear in chronological order under their respective Answers. Currently they appear in a somewhat random and changing order when their are several Replies. This has been a problem since the last "upgrade" to MaplePrimes.

Another persistent problem since that upgrade is that about half of the time that someone tries to attach a file, the file does not actually attach, even though the editor gives every indication that it has attached, and a hyperlink even appears in the post.

People are starting to leave MaplePrimes because of these problems.

Thank you for your attention to these matters,
Carl Love

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